Lemma 15.66.10. Let $A \to B$ be a ring map. Let $K^\bullet $ and $L^\bullet $ be complexes of $B$-modules. Let $a, b, c, d \in \mathbf{Z}$. If

$K^\bullet $ as a complex of $B$-modules has tor amplitude in $[a, b]$,

$L^\bullet $ as a complex of $A$-modules has tor amplitude in $[c, d]$,

then $K^\bullet \otimes ^\mathbf {L}_ B L^\bullet $ as a complex of $A$-modules has tor amplitude in $[a + c, b + d]$.

**Proof.**
We may assume that $K^\bullet $ is a complex of flat $B$-modules with $K^ i = 0$ for $i \not\in [a, b]$, see Lemma 15.66.3. Let $M$ be an $A$-module. Choose a free resolution $F^\bullet \to M$. Then

\[ (K^\bullet \otimes _ B^\mathbf {L} L^\bullet ) \otimes _ A^{\mathbf{L}} M = \text{Tot}(\text{Tot}(K^\bullet \otimes _ B L^\bullet ) \otimes _ A F^\bullet ) = \text{Tot}(K^\bullet \otimes _ B \text{Tot}(L^\bullet \otimes _ A F^\bullet )) \]

see Homology, Remark 12.18.4 for the second equality. By assumption (2) the complex $\text{Tot}(L^\bullet \otimes _ A F^\bullet )$ has nonzero cohomology only in degrees $[c, d]$. Hence the spectral sequence of Homology, Lemma 12.25.1 for the double complex $K^\bullet \otimes _ B \text{Tot}(L^\bullet \otimes _ A F^\bullet )$ proves that $(K^\bullet \otimes _ B^\mathbf {L} L^\bullet ) \otimes _ A^{\mathbf{L}} M$ has nonzero cohomology only in degrees $[a + c, b + d]$.
$\square$

## Comments (2)

Comment #5529 by Joel Stapleton on

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